Question

how to intergrate in parts
e^-(nx) sin(2nπx)

Answer 1

do the sin bit twice

Answer 2

i know but to subtitute in the interval from o to 1

Answer 3

we choose u and v and differnatite u and integrate v

Answer 4

yes

with an integral like that, it doesn't really matter which way round you do it. the exponential just adds constants, and the trig will just flip back and forward. so i'd say do it both ways for the practise

and i might whack it out as the Imaginary part of a complex number integration, but that too is way over-rated as an approach! soz

Answer 5

i'm sorry, you do know how to integrate by parts? i assumed that.

Answer 6

Ok, you're not responding but it could be my totally pitiful bandwith, and i have to go; so we can kick-off by agreeing that your integral is

\(I = \int e^{-nx} ~ \sin (2 n \pi x) ~ dx\)

re-writing the product rule \((uv)' = u' v + u v'\) as \(\int u v' = uv - \int u' v\) and then setting , willy-nilly, \(u = e^{-nx}, v' = \sin (2 n \pi x)\) means that \(u' = -ne^{-nx}, v = -\dfrac{1}{2 n \pi}\cos (2 n \pi x)\)

so we have

\(I = e^{-nx}. -\dfrac{1}{2 n \pi}\cos (2 n \pi x) - \int -\dfrac{1}{2 n \pi}\cos (2 n \pi x). -ne^{-nx} ~ dx\)

work that through and do it one more time....